Solving Classical Field Equations - uni-muenchen.de

1 Solving Classical Field EquationsRobert C. at M unchenFeynman graphs are often thought of as tools for computations in perturbative quan-tum Field theories. However, there is nothing particularly quantum about them and, infact, Feynman rules for tree diagrams also arise when one solves Classical Field Equations ofinteracting theories ( non-linear PDEs) perturbatively. I explained this in my lectureson Introduction to Quantum Field Theory and since I am not aware of a textbook treat-ment of this material (although all this is pretty standard and known to all practitioners)I decided to write up these lecture The Klein-Gordon equation for the free fieldWe start with the Klein-Gordon equation(+m2) = arises from the actionS= ddx (+m2) .The plane wave Ansatz (t,~x) = exp(i( t ~k ~x)) is a solution provided the dispersionrelation 2 ~k2=m2holds.

2 Upon the identification 7 Eand~k7 ~pthis is nothing but the relativisticdispersion relationE2=m2+~p2of a particle of rest massmin our units wherec= h= 1. As the Klein-Gordon equation islinear there is a superposition principle and any sum or multiple of solutions yields a newsolution. We use this when we write the general solution in terms of its Fourier modeswhich are plane wave solutions: (t,~x) = d d~k ( ,~k) ( 2 ~k2 m2)ei( t ~k ~x)= d~k2 (~k)(a(~k)ei( (~k)t ~k ~x)+a (~k)ei( (~k)t ~k ~x)),where we defined (~k) = ~k2+m2. The measure factor in the second line arises from the -function transformation ( (f(x)) = (x x0)/|f (x0)|forf(x0) = 0) which contributesat = (~k).12. The non-linear Field equationUltimately, we would like to solve Equations of interacting fields stemming from a generalpotentialV( ):(+m2) =V ( )For example, we are interested in monomialsV( ) =g1n n.

3 For oddn, the potential isunbounded from below and one would be able to gain in infinite amount of energy bymaking very large (or very negative). Such a system would be unstable and thus werestrict our attention to evenn. Forn= 0, the right hand side 2 gives justanother mass term and we can deal with it by 4 is the first interestingcase and we will start out studying this 4-theory .The equation (+m2) =g 3is non-linear and there is no longer a superpositionprinciple. This makes it very hard (and in general impossible) to write down exact solutionsto this equation beyond the trivial = 0 (the appendix, however, discusses a special classof exact solutions, the kinks or domain-walls ).Thus, in general, we will have to resort to an approximate procedure: Forg 1, wecan treat the right hand side as a small perturbation of the free Klein-Gordon equation andobtain approximations to the true solution by perturbing solutions to the The inhomogeneous Field equation and Green s functionsTo this end, let us step back a bit and first analyse the inhomogeneous Klein-Gordonequation(+m2) (x) =u(x)where in the right hand side we have an arbitrary function which we think of as givenrather than the RHS of 4-theory which is a function of the unknown.

4 Already thisequation is not linear but the space of solutions is still affine: When we have two solutionsto two such Equations (+m2) 1=u1and (+m2) 2=u2we obtain a solution for theequation where the RHS s add:(+m2)( 1+ 2) = (u1+u2).We can use this idea to solve this equation in the case where the RHS can be decomposedinto a sum of elementary functions:u(x) = i iui(x)If we can solve these elementary Equations (+m2) i=ui, we obtain a solution as = i i can take this idea to the extreme by decomposing the functionu(x) into a sum of -functions:u(x) = dy u(y) (x y)2 Here,u(y) plays the role of the coefficients iand the onlyxdependent function on theright hand side is the -function. So, once we have a solution to(+m2) g(x) = (x)we have a solution (x) = dy u(y) g(x y)as we can directly compute:(+m2) = (+m2) dyu(y) g(x y)= dy u(y)(+m2) g(x y)= dy u(y) (x y)=u(x).

5 This trick to solve inhomogeneous Equations is obviously not restricted to the Klein-Gordonequation. Such a solution gwhich solves an equation with a -inhomogeneity is called Green s function or propagator (in physics circles) and fundamental solution (bymathematicians). Strictly speaking, gis not really a function but in general a distribution(like the distribution) and and is obtained as a convolution which then yields a properfunction ifuis nice enough but we will not analyse this in more remains to find a solution gof (+m2) g= . As the differential operator istranslation invariant (it does not containxdependent coefficients), this can be done usinga Fourier decomposition as differentiation becomes a simplemultiplication in momentumspace. We take the Fourier transform on both sides of the equation dk 2 d((+m2) g(x))eikx= dk 2 d (x)eikx dk 2 d( k2+m2) g(x)eikx=1 2 dThis allows us to read off g(k) =1 2 d1 k2+m2for the Fourier transform gof g: g(x) = dk 2 d g(k)e ikx= dk(2 )de ikx k2+ Solving the interacting theory perturbativelyArmed with this ability to solve arbitrary inhomogeneous Equations we now come back tothe 4-equation(+m2) =g want to view this as a family of Equations parametrised by the coupling , the solutions to all these Equations will dependong.

6 Underlying the idea ofperturbation theory is the idea that these solutions can be written as a power-series ing, that they are analytic ingaroundg= , this is not really the case as can be seen as follows: Power series (inthe complex plane) have a radius of convergence (which can bezero or infinite): Everywhereinside a circle of this radius the power series converges andoutside it diverges. Thus, if thepower-series would converge for anyg >0 it would as well have to converge for someg < forg <0, again, the potential is unbounded from below and the system is unstable:Solutions will be radically different from solutions of the free equation and not be smallperturbations. In fact, as is shown in the appendix, the kinksolutions have energy andaction scaling like 1/gwhich has a singularity atg= 0.

7 In a path integral (which in astationary phase approximation reproduces the Classical behavior), these solutions appearas saddle points contributinge S e 1/g. These contributions are exponentially smallfor smallg. In fact, this function has an essential singularity atg= 0 and is invisible in aTaylor expansion around this point. Indeed, solitoninc solutions like the kink are believedto be what is missed by the perturbative treatment. Their contributions are exponentiallysmall for smallgand can thus be safely ignored if one is interested in solutions to a , we will just proceed and pretend that solutions to the 4-equation canbe written as a power series = n=0 ngnfor some coefficient functions n(x).Now plug this Ansatz into the equation and collect powers ofg: n=0(+m2) ngn=g( n=0 ngn)3= n=0( k,l,mk+l+m+1=n k l m)gnComparing coefficients we find(+m2) n= k,l,mk+l+m+1=n k l simple manipulation has helped us a lot: We can now work our ways up starting fromn= 0 to largern.

8 The important observation here is that this is a differential equationfor nin terms of a right-hand side given in terms of k, l, and mwhere allk,l,m < is, when computing nwe already know these k, l, and m!4 Let s see how this works out for the first couple ofn:(+m2) 0= 0 Nothing to be done. We know the solution is given in terms of plane waves obeying thedispersion relation. Next is(+m2) 1= 30 That was simple. Using the Green s function, we can write down the solution: 1(x) = dy g(x y) 0(y) comes(+m2) 2= 3 20 3 arises as there are three possible assignments of two 0 s and one 1 to (k,l,m). Thesolution is 2(x) = 3 dy g(x y) 0(y)2 1(y)= 3 dy dy g(x y) 0(y)2 g(y y ) 0(y )3 Now forn= 3:(+m2) 3= 3 20 2+ 3 0 iterated solution gets longer and longer: 3(x) = dy g(x y)(3 0(y)2 2(y) + 3 0(y) 1(y))= 9 dy dy dy g(x y) 0(y)2 g(y y ) 0(y )2 g(y y ) 0(y )3+ 3 dy dy dy g(x y) 0(y) g(y y ) 0(y )3 g(y y ) 0(y )35.

9 Feynman graphs in position spaceObviously, continuing like this will be more and more cumbersome. However, we see asimple pattern of these terms emerging: We can represent thesolution for 1like this:xy g 0 0 05We obtain the solution by bringing together three 0 s at one pointyand then transportthis toxusing the Green s function g. At higher orders, this pattern is iterated. Forn= 2, we haveyy 3where the factor 3 arises because the graph for 1can be substituted at any of the threelegs. At leveln= 3, there are two different graphs9yy y + 3yy y again with symmetry factors indicating the number of possibilities of obtaining this graphical notation, it should be clear what we have todo to obtain theexpression for n: We have to draw all possible graphs according to these rules: Drawnvertices for the expression for nat ordergn.

10 Each vertex gets one in-going line at the left and three outgoing lines to the right. A line can either connect to the in-going port of another vertex or to the right-hand sideof the diagram. Write down an integral for the point of each vertex. For a line connecting two vertices at pointsy1andy2, write down a Green s function g(y1 y2). For a line ending on the right. write down a factor of 0evaluated at the point of thevertex at the left of the line. Multiply by the number of permutations of outgoing lines at the vertices which yielddifferent diagrams ( symmetry factor ).66. More general Field equationsLooking back at how these rules came up, we can immediately guess the generalisation toother Field Equations : The fourth order potentialV( ) =g4 4resulted in a Field equationwith a cubic right-hand side.